92. Reverse Linked List II【Medium】

92. Reverse Linked List II【Medium】

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

解法一：

class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        if (head == NULL || head->next == NULL) {
            return head;
        }
        
        ListNode * dummy = new ListNode(INT_MIN);
        dummy->next = head;
        ListNode * mth_prev = findkth(dummy, m - 1);
        ListNode * mth = mth_prev->next;
        ListNode * nth = findkth(dummy, n);
        ListNode * nth_next = nth->next;
        nth->next = NULL;
        
        reverseList(mth);
        
        mth_prev->next = nth;
        mth->next = nth_next;
        
        return dummy->next;
    }
    
    ListNode *findkth(ListNode *head, int k) 
    {
        for (int i = 0; i < k; i++) {
            if (head == NULL) {
                return NULL;
            }
            head = head->next;
        }
        return head;
    }
    
    ListNode * reverseList(ListNode * head)
    {
        ListNode * pre = NULL;
        while (head != NULL) {
            ListNode * next = head->next;
            head->next = pre;
            pre = head;
            head = next;
        }
        
        return pre;
    }
};

 解法二：

class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        if (head == NULL || head->next == NULL) {
            return head;
        }
        
        ListNode * dummy = new ListNode(INT_MIN);
        dummy->next = head;
        head = dummy;
        
        for (int i = 1; i < m; ++i) {
            if (head == NULL) {
                return NULL;
            }
            head = head->next;
        }
        
        ListNode * premNode = head;
        ListNode * mNode = head->next;
        ListNode * nNode = mNode;
        ListNode * postnNode = mNode->next;
        
        for (int i = m; i < n; ++i) {
            if (postnNode == NULL) {
                return NULL;
            }
            ListNode * temp = postnNode->next;
            postnNode->next = nNode;
            nNode = postnNode;
            postnNode = temp;
        }
        
        mNode->next = postnNode;
        premNode->next = nNode;
        
        return dummy->next;
    }
};

解法三：

public ListNode reverseBetween(ListNode head, int m, int n) {
    if(head == null) return null;
    ListNode dummy = new ListNode(0); // create a dummy node to mark the head of this list
    dummy.next = head;
    ListNode pre = dummy; // make a pointer pre as a marker for the node before reversing
    for(int i = 0; i<m-1; i++) pre = pre.next;
    
    ListNode start = pre.next; // a pointer to the beginning of a sub-list that will be reversed
    ListNode then = start.next; // a pointer to a node that will be reversed
    
    // 1 - 2 -3 - 4 - 5 ; m=2; n =4 ---> pre = 1, start = 2, then = 3
    // dummy-> 1 -> 2 -> 3 -> 4 -> 5
    
    for(int i=0; i<n-m; i++)
    {
        start.next = then.next;
        then.next = pre.next;
        pre.next = then;
        then = start.next;
    }
    
    // first reversing : dummy->1 - 3 - 2 - 4 - 5; pre = 1, start = 2, then = 4
    // second reversing: dummy->1 - 4 - 3 - 2 - 5; pre = 1, start = 2, then = 5 (finish)
    
    return dummy.next;
    
}

参考了@ardyadipta 的代码，Simply just reverse the list along the way using 4 pointers: dummy, pre, start, then